Prove that the minimum is less than one

Question

Let $2\leq x\leq 3$ for the function :

$$f(x)=x^{(\Gamma(x+\frac{1}{2}))} (\Gamma(x+\frac{1}{2}))^{(-x)}$$

Then denotes by $x_{min}$ the abscissa of the minimum .

Prove that :

$$f(x_{min})<1$$

I have tried to attack this problem with Taylor's series and we have at $x=3$ :

$$f(x)\simeq 1.04933 + 0.655881 (x - 3) + 2.58005 (x - 3)^2 + 3.62558 (x - 3)^3 + 5.4263 (x - 3)^4 + 7.87526 (x - 3)^5 + O((x - 3)^6)$$

Now the problem is the rest that I cannot cancel easily.

Another approach is the Legendre duplication formula wich states :

$${\displaystyle \Gamma (z)\Gamma \left(z+{\tfrac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).}$$

But I think it complicate the problem more than it solves .

How to prove it ?

Any suggestions is very welcome .

Answer 1

Let us prove that there exists $x_0 \in (2, 3)$ such that $f(x_0) < 1$, or $$\Gamma(x_0 + 1/2) \ln x_0 - x_0 \ln \Gamma(x_0 + 1/2) < 0$$ or $$\frac{\ln x_0}{x_0} < \frac{\ln \Gamma(x_0 + 1/2)}{\Gamma(x_0 + 1/2)}.$$

Let $g(u) = \frac{\ln u}{u}$. $g(u)$ is strictly increasing on $(0, \mathrm{e})$, and strictly decreasing on $(\mathrm{e}, \infty)$.

Thus, it suffices to prove that there exists $x_0 \in (2, 3)$ such that $\Gamma(x_0 + 1/2) = \mathrm{e}$ which is true since $\Gamma(5/2 + 1/2) = 2$ and $\Gamma(3 + 1/2) = \frac{15}{8}\sqrt{\pi} > 3$.

We are done.

Answer 2

The derivative of the function $$f(x)=x^{\Gamma \left(x+\frac{1}{2}\right)} \Gamma \left(x+\frac{1}{2}\right)^{-x}$$ cancels very close to $$x_*=\sqrt[4]{6} \sqrt{\sqrt[3]{3}+\sqrt[3]{6}} =2.825560791\implies f(x_*)=0.9981754651$$

A rigorous minimization gives $$x_*=2.825560845 \qquad \text{and} \qquad f(x_*)=0.9981754650$$

Using a formal Taylor series to $O((x-3)^3)$ for $\log(f(x)$ (have a look here) to get for its derivative (I shall not write the formula - too long) $$x_*=2.86192\qquad \text{and} \qquad f(x_*)=1.000052879$$ This is not sufficient. Push the expansion to $O((x-3)^4)$ $$x_*=2.818051838\qquad \text{and} \qquad f(x_*)=0.9982518162$$